# wave optics (xiii)teaching notes

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Page # 1

WAVE OPTICS

Coherent and Incoherent SourcesWhy do we not commonly see interference effects with visible light? With light from a source such as theSun, an incandescent bulb, or a fluorescent bulb, we do not see regions of constructive and destructiveinterference; rather, the intensity at any point is the sum of the intensities due to the individual waves.Light from anyone of these sources is, at the atomic level, emitted by a vast number of independentsources. Waves from independent sources are incoherent; they do not maintain a fixed phase relation-ship with each other. We cannot accurately predict the phase (for instance, whether the wave is at amaximum or at a zero) at one point given the phase at another point. Incoherent waves have rapidlyfluctuating phase relationships. The result is an averaging out of interference effects, so that the totalintensity (or power per unit area) is just the sum of the intensities of the individual waves. Only thesuperposition of coherent waves produces interference. Coherent waves must be locked in with a fixedphase relationship. Coherent and incoherent waves are idealized extremes; all real waves fall some-where between the extremes. The light emitted by a laser can be highly coherent-two points in the beamcan be coherent even if sep arated by as much as several kilometers. Light from a distant point source(such as a star other than the Sun) has some degree of coherence.

Thomas Young (1773-1829) performed the first visible-light interference experiments using a clever technique toobtain two coherent light sources from a single source. When a single narrow slit is illuminated, the lightwave that passes through the slit diffracts or spreads out. The single slit acts as a single coherent sourceto illuminate two other slits. These two other slits then act as sources of coherent light for interference.

INTENSITY OF TWO SOURCE INTERFERENCEWe now obtain an expression for the distribution of intensity of two coherent sources that are in phase.The wave function in this case is the electric field. We assume that the slits are narrow enough fordiffraction to spread light from each slit uniformly over the screen. Thus, the amplitude of the fields anypoint on the screen will be equal. At a given point of the screen the fields due to S1 and S2 are

E1 = E0 win (t) ; E2 = E0 sin (t + )where the phase difference depends on the path difference = r2 r1. Since one wavelength corresponding to a phase change of 2, a distance corresponds to a phase change given by / 2 = / . If the screen is far from the slits, d sin (see the ref. figure for YDSE), therefore

= 2

=

sind2

The resultant field is found from the principle of superposition :E = E1 + E2 = E0 sin (t) + E0 sin (t + )

TEACHING NOTESTEACHING NOTES

Page # 2

By using the trigonometric identity sin A + sin B = 2 sin [(A + B)/2] cos [(A B)/2], we obtain

E = 2E0 cos

2 sin

2t

The amplitude of the resultant wave is 2E0 cos (/ 2). The intensity of a wave is proportional to thesquare of the amplitude, so from equation wave have

I = 4I0 cos2 cos2

2where I0 E0

2 is the intensity due to a single source. This function is plotted in fig. of pt (iv). The maximaoccur when = 0, 2, 4, ... = 2m. At these points I = 4I0 ; that is the intensity is four times that of asingle source. The minima (I = 0) occur when = , 3, 5, ....... = (2m + 1).

(i) If amplitudes of waves arriving at point P on the screen are different then resultant intensity is given by

I = I1 + I2 + 21II2 cos

Also, Imax = 221 II , when cos = 1Imin = 221 II , when cos = 1

(ii)min

max

II

=

2

21

21

IIII

=

2

21

21

AAAA

= 2

1r1r

where r = 2

1

2

1

II

AA

(iii) The phenomenon of interference is based on conservation of energy. There is no destruction of energy inthe interference phenomenon. The energy which apparently disappears at the minima, has actually beentransferred to the maxima where the intensity is greater than that produced by the two beams actingseparately.

Iav = 21

2

021212

0

2

0 IIcosII2II(21

d

Id

2

0

0dcos

as the average value of intensity is equal to the sum of individual intensities, therefore the energy is notdestroyed but merely redistributed in the interference pattern.

Page # 3

(iv) All maxima are equally spaced and equally bright. This is true for minima as well. Also interferencemaxima and minima are alternate. The intensity distribution in interference pattern is shown in figure.

O /2 In

tens

ity (I

) 221max III

21av III 2

21min III

Path difference ( x)

(iv) Path difference (x) and phase difference () are related as given below : path difference = 2 phase difference

or x = 2

YOUNGS DOUBLE SLIT EXPERIMENT (YDSE)The experiment set up for Youngs double slit experiment is shown in figure. Light after passing througha pin hole S is allowed to fall on thin slits S1 and S2 placed symmetrically w.r.t. S. A screen isplaced at a distance D from S1 and S2.

y

Q O

D

S1

S2

P

d

S1

S2x

MonochromaticSource Screen

Geometric construction for describing Youngs double-slit experiment

S

(D >> d)

Let P be the point, at which we want to investigate the intensity. Two rays S1P and S2P starting from S1and S2 reach P and interfere with each other.If x is the path difference between two rays,

x = S2P S1P d sin Ddy

DytansinanglesmallFor

This equation assumes that S1P and S2P are parallel, which is approximately true because D is muchgreater than d.

(a) Maxima : Point P will be a bright spot if the path difference x is integral multiple of .

yn = dnD

where, n = 0, 1, 2, 3, .......

Thus, bright spots are obtained at distances, 0, dD

, dD2

, dD3

....... from O.

(b) Minima : Point P will be a dark spot if the path difference x is an odd multiple of 2

.

Page # 4

i.e., if Dyd

= 2)1n2(

yn = d2D)1n2(

where, n = 0, 1, 2, 3, ........

Thus, dark spots are obtained at distances, d2D5,

d2D3,

d2D

....... from O.

(c) Fringe width ()It is the distance between two consecutive bright or dark fringes.

Let yn and yn 1 respectively, be the distances of nth and (n 1)th bright fringe from O,

= [yn yn1] = n dD

(n 1) dD

= dD

(n n + 1)

or = dD

Similarly, it can be proved that distance between two consecutive dark fringes, is given by

= dD

= = D / dHence, the bright and dark fringes are equally spaced.

[Home Work : HCV : 1 to 11Sheet Ex.1 : 1 Ex. 3 : 1, 6Q. Bank : 1, 3 AR : 2 ]

Remarks :(i) If whole apparatus is immersed in liquid of refractive index then,

= dD

i.e., fringes width decreases

(ii) Some times in numerical problems, angular fringe width () is given which is defined as angular separationbetween two consecutive maxima or minima

= dD

In medium, other than air or vacuum,

= d

Page # 5

(iii) x = Dyd

is valid when angular position of maxima or minima is less than 6

. However x = d sin is

valid for larger values of provided d

Page # 6

SHAPE OF INTERFERENCE FRINGES IN YDSEWe discuss the shape of fringes when two pinholes are used instead of the two slits in YDSE.Fringes are locus of points which move in such a way that its path difference from the two slits remainsconstant.

y

S1

S2

O x

Px

S2P S1P = D = constant ........(1)

If = 2

, the fringe represents 1st minima.

S2

S1

Y

X

= 3 = 2 =

= = =

If = 23

it represents 2nd minima

If = 0 it represents central maxima,If = , it represents 1st maxima etc.Equation (1) represents a hyperbola with its two foci at S1 and S2The interference pattern which we get on screen is the section of hyperboloid of revolution when werevolve the hyperbola about the axis S1S2.

(A) If the screen is to the X-axis, i.e. in the YZ plane, as is generally the case, fringes are hyperbolic witha straight central section.

(B) If the screen is in the XY plane, again fringes are hyperbolic.(C) If screen is to Y-axis (along S1S2) i.e. in the XZ plane, fringes are concentric circles with center on the

axis S1S2 ; the central fringe is bright if S1S2 = n and dark if S1S2 = (2n 1) 2

.

Z

A

Y

Y

B

X

X

Y

C[Home Work :HCV : 29 to 32Sheet Ex.1 : 2 to 6 Ex.2 : 1 Ex.3 : 7, 12, 13Q Bank MCQ : 6, 7, 8, 29 ]

{tell students to attempt from the book or can be given as HW}HCV-Ex. 33

Page # 7

Ex. Consider the situation shown in figure. The two slits S1 and S2 placed symmetrically around the centralline are illuminated by a monochromatic light of wavelength . The separation between the slits is d. Thelight transmitted by the slits falls on a screen 1 placed at a distance D from the slits. The slit S3 is at thecentral line and the slits S4 is at a distance z from S3. Another screen 2 is placed a further distance Daway from 1. Find the ratio of the maximum to minimum intensity observed on 2 if z is equal to

S1

S2

S4

S3

1

z

2D D

d

(a) d2D

(b) d4D

[Sol.(a) Let I is intensity due to slits S1 and S2 on screen S1. Further, intensity at any point on screen 1 is givenby

IP = 4I cos2

2At slit S3, = 0

3SI = 4I

At slit S4, x